# Rate of Change of Vector in Rotating Frame

• ccrook
In summary, the equation for the rate of change of a vector in an inertial frame can be expressed as: $$\Big(\frac{dQ}{dt}\Big)_{S_{0}}= \Big(\frac{dQ}{dt}\Big)_{S}+\Omega \times Q$$ where Q is an arbitrary vector. If Q is a position vector, it's usually measured from a defined origin (in this case probably inertial frame S).
ccrook
I recognize the rate of change of a vector in an inertial frame S can be related to the rate of change of the vector in a rotating frame S0 by the equation below taken from my textbook, where Ω is the angular velocity vector. $$\Big(\frac{dQ}{dt}\Big)_{S_{0}}= \Big(\frac{dQ}{dt}\Big)_{S} + \Omega \times Q$$
However, I am unsure which frame of reference Q refers to in the cross product.

Define Q.

TJGilb said:
Define Q.
Q is an arbitrary vector such as a point of a ball in space.

If Q is a position vector, it's usually measured from a defined origin (in this case probably inertial frame S). Your textbook may or may not have a picture illustrating this.

TJGilb said:
If Q is a position vector, it's usually measured from a defined origin (in this case probably inertial frame S). Your textbook may or may not have a picture illustrating this.
The origins of both frames are coincident and Q can be described in either S or S0, but the textbook doesn't clarify which frame Q is described in for the cross product.

All equations are always in one frame or, even less confusing, you work with invariant objects, namely the vectors themselves, and not with their components. In this case, let's take ##\vec{e}_j## as the time-independent basis vectors in an inertial reference frame and ##\vec{e}_k'## the time-dependent vectors in the rotating frame. Then there's a rotation matrix ##D_{kj}## such that
$$\vec{e}_k'=D_{kj} \vec{e}_j.$$
Now take any vector ##\vec{A}##. You have
$$\frac{\mathrm{d}}{\mathrm{d} t} \vec{A}=\dot{\vec{A}}=\frac{\mathrm{d}}{\mathrm{d} t} (A_k' \vec{e}_k').$$
Now you have
$$\dot{\vec{A}}=\dot{A}_k' \vec{e}_k' + A_k' \dot{\vec{e}}_k'. \qquad (1)$$
But now
$$\dot{\vec{e}}_k'=\dot{D}_{kj} \vec{e}_j=\dot{D}_{kj} (D^{-1})_{jl} \vec{e}_l'=\epsilon_{klm} \Omega_m' \vec{e}_l', \qquad (2)$$
where I have used that because of
$$\hat{D} \hat{D}^T=\hat{1}$$
the matrix ##\dot{D}_{kj} (D^{-1})_{jl}## is antisymmetric (prove it!). Plugging this in (2) in (1) finally gives
$$\dot{\vec{A}}=\dot{A}_k' \vec{e}_k' + \epsilon_{klm} A_k' \Omega_m' \vec{e}_l = (\dot{A}_l'+\epsilon_{lmk} \Omega_m' A_k') \vec{e}_l'.$$
This means that for the components of ##\dot{\vec{A}}## you get
$$\mathrm{D}_t \underline{A}'=\mathrm{d}_t \underline{A}' + \underline{\Omega}' \times \underline{A}',$$
where the underlined symbols mean the component-column vectors wrt. the basis ##\vec{e}_k'##.

ccrook said:
The origins of both frames are coincident and Q can be described in either S or S0, but the textbook doesn't clarify which frame Q is described in for the cross product.
Choose some fairly easy case, such as a unit vector along the x-axis in the ##S_0## frame, and try it both ways... See which interpretation of the formula gives you a sensible result.

I'm not suggesting this because it's the best way of resolving the ambiguity in the textbook, but because the exercise will go a long ways towards building your intuition around how these transformations work.

Chestermiller

## 1. What is the definition of "rate of change of vector in rotating frame"?

The rate of change of vector in rotating frame refers to the change in direction and magnitude of a vector in a rotating coordinate system. It is a measure of how quickly a vector is rotating or changing in a rotating frame of reference.

## 2. How is the rate of change of vector in rotating frame calculated?

The rate of change of vector in rotating frame can be calculated using vector calculus equations, specifically the formula for angular velocity. It involves taking the derivative of the vector with respect to time and accounting for the rotation of the coordinate system.

## 3. What is the significance of calculating the rate of change of vector in rotating frame?

Calculating the rate of change of vector in rotating frame is important in many fields, such as physics, engineering, and navigation. It allows us to understand the motion of objects in rotating systems and make accurate predictions and calculations.

## 4. How does the rate of change of vector in rotating frame affect the motion of objects?

The rate of change of vector in rotating frame determines the direction and speed at which an object is rotating. It also affects the forces acting on the object and can cause changes in its motion, such as precession or nutation.

## 5. Can the rate of change of vector in rotating frame be negative?

Yes, the rate of change of vector in rotating frame can be negative. This indicates that the vector is rotating in the opposite direction of the rotating frame or that its magnitude is decreasing over time. It is a common occurrence in rotational motion and does not necessarily indicate a problem or error.

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